Generating And Saving An .eml File With Python 3.3
Solution 1:
You are supposed to pass an open file (in write mode) to Generator()
. Currently you pass it just a string, which is why it fails when it tries to call .write()
on the string.
So do something like this:
import os
cwd = os.getcwd()
outfile_name = os.path.join(cwd, 'message.eml')
classGen_Emails(object):
# ...defSaveToFile(self,msg):
withopen(outfile_name, 'w') as outfile:
gen = generator.Generator(outfile)
gen.flatten(msg)
Note: with open(outfile_name, 'w') as outfile
opens the file at the path outfile_name
in write mode and assigns the file pointer to the open file to outfile
. The context manager also takes care of closing the file for you after you exit the with
block.
os.path.join()
will join paths in a cross-plattform way, which is why you should prefer it over concatenating paths by hand.
os.getcwd()
will return your current working directory. If you want to your file to be saved somewhere else just change it out accordingly.
Solution 2:
Here is a modified solution that works with extra headers too. (This was tested with Python 2.6)
import os
from email import generator
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText
html_data = ...
msg = MIMEMultipart('alternative')
msg['Subject'] = ...
msg['From'] = ...
msg['To'] = ...
msg['Cc'] = ...
msg['Bcc'] = ...
headers = ... dict of header key / value pairs ...
for key in headers:
value = headers[key]
if value andnotisinstance(value, basestring):
value = str(value)
msg[key] = value
part = MIMEText(html_data, 'html')
msg.attach(part)
outfile_name = os.path.join("/", "temp", "email_sample.eml")
withopen(outfile_name, 'w') as outfile:
gen = generator.Generator(outfile)
gen.flatten(msg)
print"=========== DONE ============"
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